Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Today I was fiddling about with a TI-89 calculator, attempting as usual to confuse it. I figured that making it solve an equation with a periodic function would be fun, so I tried the following:

$$\cos(\pi x) + x^2 = 0$$

While this didn't stump it, I noted that the solution it gave was just a decimal. Since I ran this through solve(), which usually will give you a nice solution like $\frac{\sqrt{2}}{2}$ if it can, I found this rather interesting. I figured what it eventually did was give up on solving it in its usual manner and move to a numeric method: i.e., it simply made intelligent guesses until it found good solutions.

I am usually inclined to think most random decimals have a closed form expression behind them. (In fact, do all such decimals have a closed form, even if we don't know it? I may have to look into that.) As such, I decided to put this into Wolfram|Alpha and see if it had any better results. But no dice; it gave me back, similar to the TI-89, that $x = \pm 1$ and $x \approx \pm 0.629847$ were solutions.

The latter decimal, $x \approx 0.629847$, is the one I am concerned with. As far as I am concerned, $x = \pm 1$ are sort of "trivial" solutions; just thinking the problem over leads you to them naturally.

Is there a way to solve this algebraically? I can sort of narrow it down. I know that as $x \to \infty$, the $\cos(\pi x)$ term is essentially trivial compared to $x^2$. Given that $x^2>0$ for any $x \ne 0$, and given that $\cos(\pi x)$ has a range of $[-1, 1]$, it seems to me that whenever $x^2>1$, $\cos(\pi x)$ cannot pull down $x^2$ enough for it become zero.

So, it seems natural to me to think, then, that all solutions must lie where $x^2 \le 1$, viz., within the interval $[-1, 1]$. This narrows down the field significantly, but it still does not really help me with an algebraic solution. (However, I figure if I were to come across this in a real-world scenario, this would be a useful line of attack for a guess-and-test sort of deal.)

Another line of attack I attempted was to take the reverse approach: use the numerically attained solution to find a closed-form solution. I thought it may be an interesting number I simply had not learned about, so I tried to look up the decimal sequence in the OEIS, but to no luck: no such sequence was available.

I've sort of rambled, so here are my questions:

  1. Is there an algebraic solution to the above equation?
  2. Even if there is not, is there any way to figure out the closed form expression behind the decimal $x = 0.629847$? I don't even care if the expression has $\cos$ or $\sin$ in it.

I will be honest: I really don't even know where to start.

share|cite|improve this question
(1) Highly doubtful. (2) Every finite decimal has the closed-form of a rational number, but "almost all" infinite decimals have no closed form. This has to do with cardinality: the set of all definable numbers will be countably infinite while the real numbers are uncountably many. – anon May 17 '12 at 3:25
As a nice, general, rule-of-thumb: any transcendental equation where the variable you're solving for appears within the transcendental function (in your case, the cosine) and outside the function, will generally not have a simple closed-form solution. – J. M. May 17 '12 at 3:27
Expressing cosine in terms of exponentials, can give $(-1)^{x+1} = x^2.$ (I might've made a mistake, though.) – user2468 May 17 '12 at 3:36
Closed form, in general, is a very deceptive term. What do you mean by it? If you get a solution say $\pi/4$, will you call it closed form? If so, then I can define, $\Upsilon$ is the real number number between $0$ and $1$ such that $\cos(\pi \Upsilon) + \Upsilon^2 = 0$. And now lo and behold, your $x = \Upsilon$. Would you call this closed form? It really doesn't mean much if you think about it. – user17762 May 17 '12 at 6:57
up vote 9 down vote accepted

Your $x \approx 0.629847$ is transcendental. First, $x \neq 0, \; 1/3, \; 2/3, \; 1 \;$ so by Corollary 3.12 in Niven, Irrational Numbers, page 41, $x$ is irrational. Now, one value of $$ (-1)^x $$ is $$ e^{i \pi x} = \cos \pi x + i \sin \pi x. $$ As $x$ is not rational, Gelfond-Schneider, Theorem 10.1 on page 134, $ (-1)^x $ is transcendental.

The algebraic numbers in $\mathbb C$ are a field containing $i,$ the rationals, and closed under complex conjugation. It follows that a number is algebraic if and only if both its real and imaginary parts are algebraic. From $\cos^2 \pi x + \sin^2 \pi x = 1,$ it follows that either both parts are algebraic or both are transcendental. Therefore $\cos \pi x$ is transcendental. Since $x^2 = - \cos \pi x,$ we find that $x$ itself is transcendental.

Meanwhile, lots of nice numbers are transcendental. $\pi, \; e, \; \log 2 \; $ are transcendental but would be considered pleasant answers to a problem of this type. It is very, very hard to show that a number such as your $x$ does not have a nice closed form expression. I can't see how it could, of course.

share|cite|improve this answer

The inverse symbolic calculator gives a few dozen numbers starting with .629847; maybe if you could get a couple more decimal places you'd narrow it down. Of course, Will is right that it's unlikely to have a simple expression in terms of familiar constants such as $\pi$, $\sqrt2$, etc., but it still might be a value of some special function someone has tabulated.

share|cite|improve this answer
Thanks for the tool. Unfortunately, while the results look promising for the first few digits, even expanding to the next "section" of digits with Wolfram|Alpha ends up with a decimal that's not in that tool's database. – Reid May 17 '12 at 13:41
Well, I'm not all that surprised, but I figured it was worth a try - and now you know the "calculator" is there, maybe it will come in handy in the future. – Gerry Myerson May 17 '12 at 13:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.